Question

A particle moves from A to B in a circular path of radius R covering an angle $\theta$ as shown in the figure, find the ratio of the magnitude of distance and displacement of the particle.

• A. $\frac{\theta }{2\sin \frac{\theta }{2}}$
• B. $\frac{\frac{\theta }{2}}{\sin \theta }$
• C. $\frac{\theta }{2\sin \theta }$
• D. $\frac{\theta }{\sin \frac{\theta }{2}}$

Answer 1

Answer: (A)

$\frac{\theta }{2\sin \frac{\theta }{2}}$

In triangle ABO, $\angle AOB=\theta$

AB is the displacement of the particle.

Using $\cos \theta =\frac{A{O}^2+B{O}^2-A{B}^2}{2\left(AO\right)\left(BO\right)}$

Or $\cos \theta =\frac{R^2+R^2-A{B}^2}{2R^2}$

We get $A{B}^2=2R^2-2R^{2}cos\theta$

Or $A{B}^2=2R^2(1-\cos \theta )=2R^2\times 2{\sin }^2\left(\frac{\theta }{2}\right)$

$⟹AB=2R\sin \frac{\theta }{2}$

Distance covered by the particle $d=R\theta$

Thus ratio of distance and displacement $\frac{d}{AB}=\frac{\theta }{2\sin \frac{\theta }{2}}$